Find the four numbers in A.P., whose sum is 50
Question:

Find the four numbers in A.P., whose sum is 50 and in which the greatest number is 4 times the least.

Solution:

Here, we are given that four number are in A.P., such that there sum is 50 and the greatest number is 4 times the smallest.

So, let us take the four terms as $a-d, a, a+d, a+2 d$.

Now, we are given that sum of these numbers is 50, so we get,

$(a-d)+(a)+(a+d)+(a+2 d)=50$

$a-d+a+a+d+a+2 d=50$

$4 a+2 d=50$

$2 a+d=25 \ldots \ldots$ (1)

Also, the greatest number is 4 times the smallest, so we get,

$a+2 d=4(a-d)$

$a+2 d=4 a-4 d$

$4 d+2 d=4 a-a$

$6 d=3 a$

$d=\frac{3}{6} a$………..(2)

Now, using (2) in (1), we get,

$2 a+\frac{3}{6} a=25$

$\frac{12 a+3 a}{6}=25$

$15 a=150$

$a=\frac{150}{15}$

$a=10$

Now, using the value of a in (2), we get

$d=\frac{3}{6}(10)$

$d=\frac{10}{2}$

$d=5$

So, first term is given by,

$a-d=10-5$

$=5$

Second term is given by,

$a=10$

Third term is given by,

$a+d=10+5$

$=15$

Fourth term is given by,

$a+2 d=10+(2)(5)$

$=10+10$

 

$=20$

Therefore, the four terms are $5,10,15,20$.

 

 

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